\newproblem{lay:2_1_20}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 2.1.20}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
  Suppose that the first two columns, $\mathbf{b}_1$ and $\mathbf{b}_2$, of $B$ are equal. What can be said about the columns of $AB$? Why?
}{
  % Solution
	Let us consider the different columns of $B$
	\begin{center}
	   $B=\begin{pmatrix}\mathbf{b}_1 & \mathbf{b}_2 & \mathbf{b}_3 & ... \end{pmatrix}$
	\end{center}
	The product of $AB$ is
	\begin{center}
		$AB=A\begin{pmatrix}\mathbf{b}_1 & \mathbf{b}_2 & \mathbf{b}_3 & ... \end{pmatrix}=\begin{pmatrix}A\mathbf{b}_1 & A\mathbf{b}_2 & A\mathbf{b}_3 & ... \end{pmatrix}$
	\end{center}
	If $\mathbf{b}_1=\mathbf{b}_2$, then
	\begin{center}
		$A\mathbf{b}_1=A\mathbf{b}_2$
	\end{center}
	So, both columns are also equal. Additionally, we may say that the columns of $AB$ are not linearly independent because there exists a linear combination of them
	that produces the vector $\mathbf{0}$.
	\begin{center}
		$1(A\mathbf{b}_1)-1(A\mathbf{b}_2)+0(A\mathbf{b}_3)+0(A\mathbf{b}_4)+...=\mathbf{0}$
	\end{center}
}
\useproblem{lay:2_1_20}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
